A circular current carrying wire having radius $R$ is placed in $X Y$ -plane with its centre at middle $O$ . There is a non-uniform magnetic field $B = \frac{B_{0} x}{2 R} \hat{k}$ (here $B_{0}$ is a positive constant) is existing in the region.
Given image
The magnetic force acting on semicircular wire will be along

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$- X$ -axis

$+ Y$ -axis

$- Y$ -axis

$+ X$ -axis

answer is A.

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Detailed Solution

$B = \{\begin{cases} |\frac{B_{0} x}{2 R}| \hat{k}, x > 0 \\ |\frac{B_{0} x}{2 R}| - \hat{k}, x < 0 \end{cases}$

$F_{n e t} = F_{x} \hat{i} + F_{y} \hat{j} F_{y} = Σ d F_{y} = 0 F_{x} = Σ d F_{x} = - 2 \int_{0}^{\frac{π}{2}} d F \cos θ$

Hence, $F_{n e t}$ is along negative $X$ -axis.

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